[An "application" of Existential Closedness / the Nullstellensatz]{#An_"application"of_Existential_Closedness/_the_Nullstellensatz .mw-headline}[]

When originally writing the article, I tried to include the following fact as a corollary of the existential closedness of algebraically closed fields in the class of fields:

If V is a geometrically integral variety over a field K (i.e., the base change of V to the algebraic closure of K is integral), then every base change of V is integral.

The proof ended up getting a bit out of hand. I'm moving it to this talk page since it got a bit off topic for this article:

As another "application", let us prove:

Corollary: Let K be an algebraically closed field, and L be a field extending K. Let A be a K-algebra, possibly infinite dimensional. Suppose A is integral. Then so is {\displaystyle A\otimes _{K}L}.

(In algebro-geometric terms, this means that if a variety is geometrically integral, it remains integral under any base change.)

Proof: We will instead prove a slightly stronger statement: let V and W be K-vector space, and let {\displaystyle \mu :V\otimes _{K}V\to W} be some K-bilinear pairing. Tensoring everything with L, we get

{\displaystyle \mu _{L}:(V\otimes _{K}L)\otimes _{L}(V\otimes _{K}L)\to W\otimes _{K}L}

Suppose that {\displaystyle \mu (v_{1},v_{2})\neq 0} whenever {\displaystyle v_{1}\neq 0\neq v_{2}}. Then {\displaystyle \mu _{L}} has the same property.

To see this, write V as a directed limit of finite dimensional submodules

{\displaystyle V=\bigcup _{i\in I}V_{i}}

Let {\displaystyle W_{i}\subset W} be the K-span of {\displaystyle \mu (V_{i},V_{i})}, so that {\displaystyle \mu } induces a map {\displaystyle V_{i}\otimes V_{i}\to W_{i}} for each {\displaystyle i\in I}, and the two maps

{\displaystyle V_{i}\otimes V_{i}\to W_{i}\hookrightarrow W}

{\displaystyle V_{i}\otimes V_{i}\hookrightarrow V\otimes V\to W}


The functor {\displaystyle -\otimes _{K}L} is a left adjoint, so it preserves direct limits. Also, because K is a field, L is flat, so this functor preserves injections. Therefore,

{\displaystyle V\otimes _{K}L=\bigcup _{i\in I}(V_{i}\otimes _{K}L)}

So, if there exist two non-zero vectors {\displaystyle v_{1},v_{2}\in V\otimes _{K}L} with {\displaystyle \mu _{L}(v_{1},v_{2})=0}, then some {\displaystyle V_{i}\otimes _{K}L} already contains {\displaystyle v_{1},v_{2}}.

Replacing V with Vi and W with Wi, we may assume that V and W are finite dimensional. In fact, we may assume that they are Kn and Km for some n and m. Then {\displaystyle \mu } is described by structure coefficients:

{\displaystyle \mu (e_{i},e_{j})=\sum _{k}c_{ijk}e_{k}}

where the ei are standard basis vectors. The same structure coefficients are correct after tensoring with L.

Our assumption means that in K, the following system of equations cannot be solved:

{\displaystyle \bigwedge _{k=1}^{m}\sum _{i,j}c_{ijk}v_{1}^{i}v_{2}^{j}}

{\displaystyle \prod _{i=1}^{n}v_{1}^{i}\neq 0}

{\displaystyle \prod _{i=1}^{n}v_{2}^{i}\neq 0}

in the variables {\displaystyle v_{1}^{1},v_{1}^{2},\ldots ,v_{1}^{n},v_{2}^{1},v_{2}^{2},\ldots ,v_{2}^{n}}. By existential closedness of K in L, the same system of equations cannot be solved in L, which is what we wanted to show. QED.

Will Johnson (talk) 07:15, November 15, 2013 (UTC) ::: ::: :::